From an aeroplane flying above a river at an altitude of 1200 m, it is observed that the angles of depression of opposite points on the two banks of a river are 30° and θ. If the width of the river is 3000 m, then which one of the following is correct?
45° < θ < 60°
Understanding the Problem: Aeroplane Altitude and River Width
This question involves trigonometry, specifically the concept of angles of depression and how they relate to the height of an object above the ground and the distances along the ground. We are given the altitude of an aeroplane flying above a river, the total width of the river, and one angle of depression to a point on one bank. We need to find the range of the angle of depression to a point on the opposite bank.
Setting up the Geometry
Imagine the aeroplane is at a point P, directly above a point O on the river. The altitude of the aeroplane is the vertical distance PO, which is given as 1200 m. Let A and B be points on the opposite banks of the river such that the aeroplane's position P is in the vertical plane containing A and B. The width of the river is the distance AB, given as 3000 m.
The angles of depression from the aeroplane P to the points A and B on the banks are given as 30° and θ. The angle of depression from P to A is the angle between the horizontal line through P and the line of sight PA. Similarly, the angle of depression from P to B is the angle between the horizontal line through P and the line of sight PB.
Due to the property of alternate interior angles, the angle of depression from P to A is equal to the angle of elevation from A to P (∠PAO). Similarly, the angle of depression from P to B is equal to the angle of elevation from B to P (∠PBO).
We can form two right-angled triangles: ▵POA and ▵POB, where O is the point on the river bed directly below the aeroplane (or on the line AB if A, O, B are collinear on the river surface, which is implied for width calculation).
- Altitude PO = 1200 m
- Angle of elevation from A, ∠PAO = 30°
- Angle of elevation from B, ∠PBO = θ
- Width of the river AB = 3000 m. Since A and B are on opposite banks and O is between them, AB = AO + OB.
Using Trigonometry to Relate Distances and Angles
In a right-angled triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side.
In ▵POA:
\(\tan(\angle\text{PAO}) = \frac{\text{PO}}{\text{AO}}\)
\(\tan(30^\circ) = \frac{1200}{\text{AO}}\)
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
So, \(\frac{1}{\sqrt{3}} = \frac{1200}{\text{AO}}\)
\(\text{AO} = 1200\sqrt{3}\) meters.
In ▵POB:
\(\tan(\angle\text{PBO}) = \frac{\text{PO}}{\text{OB}}\)
\(\tan(\theta) = \frac{1200}{\text{OB}}\)
So, \(\text{OB} = \frac{1200}{\tan(\theta)}\) meters.
Calculating the Unknown Angle Range
The total width of the river is the sum of the distances AO and OB:
\(\text{AB} = \text{AO} + \text{OB}\)
\(3000 = 1200\sqrt{3} + \frac{1200}{\tan(\theta)}\)
Now, we need to solve for \(\tan(\theta)\). First, subtract \(1200\sqrt{3}\) from both sides:
\(3000 - 1200\sqrt{3} = \frac{1200}{\tan(\theta)}\)
Rearrange the equation to find \(\tan(\theta)\):
\(\tan(\theta) = \frac{1200}{3000 - 1200\sqrt{3}}\)
We can factor out 1200 from the denominator:
\(\tan(\theta) = \frac{1200}{1200(2.5 - \sqrt{3})}\)
\(\tan(\theta) = \frac{1}{2.5 - \sqrt{3}}\)
To estimate the value, we use the approximate value of \(\sqrt{3} \approx 1.732\).
\(\tan(\theta) \approx \frac{1}{2.5 - 1.732}\)
\(\tan(\theta) \approx \frac{1}{0.768}\)
Now, let's calculate the approximate value:
\(\tan(\theta) \approx 1.302\) (approximately)
Now we need to find the angle \(\theta\) whose tangent is approximately 1.302. We can compare this value to the tangents of standard angles:
- \(\tan(45^\circ) = 1\)
- \(\tan(60^\circ) = \sqrt{3} \approx 1.732\)
Since \(1 < 1.302 < 1.732\), it follows that \(\tan(45^\circ) < \tan(\theta) < \tan(60^\circ)\).
Since the tangent function is increasing for angles between 0° and 90°, this inequality implies:
\(45^\circ < \theta < 60^\circ\)
This is the range for the unknown angle of depression θ.
Comparing with Given Options
Let's check the provided options:
- Option 1: \(\theta < 30^\circ\) (Incorrect, as we found \(\theta > 45^\circ\))
- Option 2: \(30^\circ < \theta < 45^\circ\) (Incorrect, as we found \(\theta > 45^\circ\))
- Option 3: \(45^\circ < \theta < 60^\circ\) (Matches our calculated range)
- Option 4: \(60^\circ < \theta < 90^\circ\) (Incorrect, as we found \(\theta < 60^\circ\))
Therefore, the correct range for the angle \(\theta\) is \(45^\circ < \theta < 60^\circ\).
| Quantity | Value |
|---|---|
| Aeroplane Altitude (PO) | 1200 m |
| River Width (AB) | 3000 m |
| Angle of depression to Bank A | 30° |
| Angle of depression to Bank B | \(\theta\) |
| Distance AO | \(1200\sqrt{3}\) m |
| Approximate \(\tan(\theta)\) | \(\approx 1.302\) |
| Range of \(\theta\) | \(45^\circ < \theta < 60^\circ\) |
Revision Table: Angles of Depression and Elevation
| Term | Definition | Relationship |
|---|---|---|
| Angle of Depression | The angle between the horizontal line of sight and the line of sight downwards to an object. | Equal to the angle of elevation from the object to the observer (alternate interior angles). |
| Angle of Elevation | The angle between the horizontal line of sight and the line of sight upwards to an object. | Equal to the angle of depression from the object to the observer. |
| Altitude | The vertical distance from an object to a reference plane (like the ground or sea level). | Forms a side in right-angled triangles used in trigonometry problems. |
Additional Information: Applying Trigonometry to Height and Distance Problems
Problems involving height and distance often require the use of trigonometric ratios (sine, cosine, and tangent) in right-angled triangles. Key steps usually involve:
- Drawing a clear diagram representing the problem.
- Identifying the right-angled triangles involved.
- Labeling the known sides and angles, and the unknown quantity to be found.
- Using the appropriate trigonometric ratio that relates the known and unknown parts of the triangle.
- Using the values of trigonometric ratios for standard angles (0°, 30°, 45°, 60°, 90°).
- Solving the resulting equation to find the unknown quantity.
In this river width problem, the altitude provides the 'height' and the distances to the banks provide the 'base' segments, linked by the tangent function and the angles of depression/elevation.
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